Talk:Lebesgue integration

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Notation[edit]

What does simple mean here? --Zslevi (talk) 20:35, 18 November 2009 (UTC)


Initial comments[edit]

Hello! I think this article is looking very good. Over the past several months many improvements have been made. I wonder if it would be better titled "Lebesgue integral", since (if I'm not mistaken) that term is rather more common in text books. Certainly "Lebesgue integral" is more common than "Lebesgue integration" on the web as shown by Google searches. Likewise "Lebesgue-Stieltjes integral" is more common on the web than "Lebesgue-Stieltjes integration". There do exist Lebesgue integral and Lebesgue-Stieltjes integral in WP but these are redirects. However, Riemann integral is an article and Riemann integration is a redirect; also integral is an article and integration is a disambiguation page (there is no integration (mathematics) article). So: in summary, I propose that we move Lebesgue integration to Lebesgue integral and likewise with Lebesgue-Stieltjes. I look forward to your comments. Regards & happy editing, Wile E. Heresiarch 16:57, 21 Jun 2004 (UTC)

On the whole, page names migrate towards the more abstract term. I think this is more suitable for WP, really: so that e.g. nerve is thought of under nervous system first (haven't checked the actual status of those). So I'd be happy to leave it as Lebesgue integration. Charles Matthews 18:08, 21 Jun 2004 (UTC)
Hmm, I guess I'm not convinced. The Lebesgue integral is the object of interest, and the purpose of Lebesgue integration is to construct the Lebesgue integral, is it not? What other items are considered under the general heading of Lebesgue integration? Perhaps I don't get out enough, of course; I wouldn't be at all surprised. Happy editing, Wile E. Heresiarch 23:31, 22 Jun 2004 (UTC)
Maybe one way would be to have an article like integration theories in mathematical analysis, as a sort of umbrella; and then call the articles on particular theories Riemann integral, Lebesgue integral and so on. Well, such an article would be useful in itself, as a survey. Charles Matthews 17:05, 23 Jun 2004 (UTC)

This seems to me like another instantiation of the "thing", "thing theory" problem. MarSch 15:20, 11 Mar 2005 (UTC)


Hello. About notation, under "Equivalent formulations" there are L1 and Cc. Do these want to be italicized as L1 and Cc or maybe Cc ? I'm just wondering how these are conventionally presented. Happy editing, Wile E. Heresiarch 14:14, 13 Apr 2004 (UTC)

I think Lang and Rudin both use L1 and Cc. Loisel 06:21, 16 Apr 2004 (UTC)

Hello, I've reworked the "Introduction" section to situate the Lebesgue integral historically, and to summarize the important differences with the Riemann integral. Then there are definitions and properties, and then there are the more detailed discussion sections. I hope this organization address the concerns with the "Rudinesque" previous revision. FWIW I'm not really happy with the "Lebesgue vs Riemann" aspect of this article, which might suggest there are exactly 2 defns of the integral; it's more "Lebesgue vs every predecessor". Also, I think "failure of the Riemann integral" gives a mistaken impression -- the Riemann integral has been a big success overall. Well, hope this helps. Wile E. Heresiarch 20:09, 8 Feb 2004 (UTC)

All I know is, after I studied the Riemann integral as an undergrad, I thought it was pretty hot stuff, so when I heard about this thing called Lebesgue integration, it took quite a bit of argument to convince me that this new-fangled thing was even necessary in the first place. The introduction is a good start, but I would take select parts of the "discussion" and merge them. As for giving the impression there are 2 defs of "integral", this is more a misunderstanding of math in general...people think there is some Platonic concept called "the integral" that mathematicians go out and "discover" its "true nature"...when there are just various definitions and theories that just are what they are. It's very difficult to explain this. Maybe it should be pointed out that the Riemann and Lebesgue integrals aren't the only theories of integration that have been developed -- just the most popular and widely used. Revolver 22:56, 8 Feb 2004 (UTC)

I don't think the discussion should come after the formal definitions and results. If someone is unfamiliar with the Lebesgue integral (and most will be who read the article) then the formal definition will seem pointless without some kind of motivation or discussion. As the article is now, someone who doesn't know the Lebesgue integral is likely to just give up part-way through the formal derivations and never reach the discussion. We're not trying to imitate Rudin. Revolver 02:06, 7 Feb 2004 (UTC)

If the definitions are incomprehensible, the discussion is irrelevant (and likewise incomprehensible). The most you can say to someone who doesn't understand the definition is "All definitions of the integral are the same for easy cases, but the Lebesgue integral handles some more difficult cases."
Well, before the Lebesgue integral was developed, many mathematicians discussed why they thought a new type of integral was necessary, what problems it should attack, what specific examples motivated it. The idea that it's impossible to "discuss" a mathematical subject without going through the details of the definitions doesn't seem right to me -- again, if for no other reason than that the mathematicians did it themselves for years before arriving the modern definition (think of groups, topological spaces, non-Euclidean geometries, etc.) I think it's quite possible to explain the specific issues that eventually led to the creation of the Lebesgue integral, without defining the Lebesgue integral itself. And yes, I'll try to do this and put it on the talk page someday...after I finish my thesis, find a job, move, etc.,...;-) Revolver 22:56, 8 Feb 2004 (UTC)

I think the (short) section just before the definition should say that. A secondary problem is that the discussion, as it stands, is needlessly verbose; it needs a lot more focus. Wile E. Heresiarch 09:20, 7 Feb 2004 (UTC)

What should ideally be done, is that the "discussion" and the "formal definition" should become merged into a single thread of narration, not separated into two pieces. This of course, would take a good deal of thought and effort to do it right. Revolver 02:08, 7 Feb 2004 (UTC)

I'm not convinced that is desirable. Maybe you can put a version of this kind of presentation on your talk page and have people look at it. Just a thought. Wile E. Heresiarch 09:20, 7 Feb 2004 (UTC)

Wile, see below, there's a guy who claims this article is too advanced. The text now under "Discussion" was my attempt at answering that gripe.

Loisel 21:13, 6 Feb 2004 (UTC)

Hello Loisel, thanks for bringing up this topic. I moved the discussion sections below the definitions, etc., because the discussion is quite long winded and goes off on several tangents. That's fine, really, since the Lebesgue integral is important and there's a lot to be said about it. But the article works better as a reference if we get right to the point.
Maybe to address the concern that the article is too abstract, we can make the "Introduction" section (just before the defn) better (not necessarily longer). Without getting technical, one can say the Lebesgue integral is more general than some other defns, and also the it's defined as the limit of a sequence of simpler integrals. Perhaps an example of a sequence of simple functions will make a good companion article. Wile E. Heresiarch 01:49, 7 Feb 2004 (UTC)

Michael, why did you remove the previous article and leave a corpse in its stead? I've now restored the article.

Loisel 08:04, 3 Feb 2004 (UTC)



I've made a correction to the Technical Difficulties regarding improper Reimann Integrals on this page. What was defined was not the Improper Reimann Integral but the improper Cauchy Pricipal Value. The improper Reimann Integral does not exist in this example.

-joshua


Look - I'm sure that this is a well-writen article, but could someone (knowledgeable) put in a 2-3 line explanation of what it actually about, and what it is used for? The opening sentence "Let m be ..." isn't really au fait in a generalist encyclopedia. I'm really glad we have this sort of advanced stuff, but in this case I don't even understand what I don't understand. - MMGB

Shouldn't it be at some point made clear that m is the Lebesgue measure? Otherwise, there's no guarantee the integral will correspond to Riemann integration - if m is the counting measure, it will correspond to summation. But I'm not sure where precisely to work that in.


The link to monotone convergence theorem is to a different theorem.

Charles Matthews 14:00, 6 Sep 2003 (UTC)

Fixed. Loisel 06:38, 7 Sep 2003 (UTC)

Too advanced![edit]

This encyclopedia article is too advanced; the only people who will be able to get the meaning are those who have already learned Lebesgue theory thoroughly. The modern, extremely compact, definitions of mathematics are simply a shorthand for people already familiar with the underlying concepts and some illustrative examples; without including the latter, the former is unintelligible.

In fairness, an article on the Lebesgue integral is certainly only of interest to a specialized audience. And it would perhaps be appropriate to include the existing article as an advanced appendix as a minimal, extremely formal, refresher, less formal explanation is needed for anyone who doesn't remember or hasn't learned what ideas the formal definition corresponds to.

For the main body of the article, it would be nice to follow history somewhat: (1) explanation of the problems encountered with the Riemann integral including specific functions or applications (2) elementary definition (NOT a formal argument: emphasis on underlying IDEAS, not on formal definitions) and explanation of how the problems of Riemann are resolved by Lebesgue (that is, examples) (3) perhaps a little modern history on the growth of analysis since Lebesgue. The formalities could be a conclusion of sorts. I would write such an article myself, but I don't know Lebesgue theory sufficiently well (hence my need to refer to an encyclopedia article); perhaps someone else might? (Anon.)

A page on improper integrals in the Riemann theory would be a help. But no one-page intro to the Lebesgue integral is going to be easy.

Charles Matthews 16:10, 23 Oct 2003 (UTC)


This is a valid point, however I'm not very good at history. I could attempt to vulgarize the concepts discussed in the "formal construction" section, but I couldn't color the text with historical anecdotes about integration theory: I don't know any.

Would that help?

Loisel 16:47, 24 Oct 2003 (UTC)

Is this better? Loisel 18:15, 24 Oct 2003 (UTC)

I'm sorry, this topic is advanced pdenapo Wed Jan 14 03:09:34 UTC 2004

Well done[edit]

With the addition of the new introductory material, this article takes the place of a good professor - at first giving a general indication of the questions and problems at hand, and once the ideas involved are explained a little further (with some nice examples) providing the formal framework which enables those ideas to be put into rigorous use. Great job.

--

Couple of points where I can carp (I do concur with the post above: the work on this page was very worthwhile).

  • The Lebesgue approach is not the most elementary area-based integration theory; that distinction goes to the Riemann integral.

Don't think that's actually true (cf. Dieudonne's Treatise on Analysis ona sub-Riemann theory).

  • Uniform convergence of Fourier series.

Rare? A couple of derivatives will do.

Charles Matthews 07:24, 18 Dec 2003 (UTC)

About the uniform convergence: it is true that twice differentiable periodic functions have uniformly convergent Fourier series. This is a very thin set in L^2. That is one of the meanings of "rare." Feel free to adjust the wording if you think you can improve it, but the goal of that passage (to show that there are many common examples where the uniform convergence theorem is insufficient) should remain, I think.

About "elementary." I meant to recognize the order in which this is usually taught. I don't know the sub-Riemann theory you mention, but I should've thought there would be a gazillion variants that claim to be more elementary than the Riemann integral. Again, feel free to adjust the wording if you think you can improve it. If you're changing it, I think it should probably say that Riemann is "more elementary" than Lebesgue, but I doubt it should refer to some obscure theory that's not generally taught.

Anyway, have a ball. Loisel 02:04, 8 Jan 2004 (UTC)

Charles Matthews refers here to what Dieudonne calls regulated functions in the English translation (satisfying a condition something like: both one sided limits exist at every point = uniform norm closure of linear span of indicator functions of intervals ). It is more elementary in the sense that all its properties follow by continuity. Dieudonne disparages theRiemann integral (which in fact is the prevalent attitude among the mathematicians I know)

I just removed the following text:

Correction: The improper Reimann integral does not exist for f or g since the improper Reimann Integral is defined as a double limit and you cannot subtract infinities, i.e. the improper Reimann integral is defined as limlim∫abf(x) dx where the limits are taken as a goes to infinity and b goes to infinity. What is true is that the Improper Cauchy Principal value (about zero) for f does in fact exist and it is PV∫-∞f(x)=0.

I will modify the text to say something like (sometimes called the Improper Cauchy Principal value about zero). Please note Wikipedia:Integrate_changes, which is part of the Wikipedia policy, requests that you integrate your changes so that they form a seamless part of the article. Loisel 02:14, 8 Jan 2004 (UTC)

I think that maybe this whole paragraph should be junked, since with the proper definition with 2 limits, there is no preferred point and no translation noninvariance. Thus this is not a deficiency of the Riemann integral.MarSch 15:30, 11 Mar 2005 (UTC)

I've rewriten this article since the old version focused only on the technical difficulties of Riemman integral, rather than defining the concept of Lebesgue integral. I've included the examples in the old version, though pdenapo Wed Jan 14 03:09:34 UTC 2004

The new introduction is nice, but not consitent with the definition that appears below. [User:pdenapo|pdenapo]]Wed Jan 14 03:09:34 UTC 2004

I think the rewrite has destroyed a large quantity of useful information and has generally decreased the quality of the article. I'm tempted to revert.

Loisel 08:05, 1 Feb 2004 (UTC)

Equivalent Formulations[edit]

I think this page is very good. Hello, Do you have references for the equivalent formulations of the integral? (unique continuous extension of a linear functional...) Thanks Sergei Vieira srgvie2000@yahoo.com.br

That's elementary functional analysis. See either Rudin's Real and Complex Analysis or Rudin's Functional Analysis, for instance. If f is a continuous functional defined on X, a dense subset of the Banach space Y, and if g and h are continuous functionals that agree with f when restricted to X, then g=h. To see this, let x be arbitrary in Y and let x_k be a Cauchy sequence in X converging to x. Then g(x_k) converges to g(x) and h(x_k) converges to h(x). But g(x_k)=h(x_k)=f(x_k), and f(x_k) is Cauchy. Hence f(x_k) has a unique limit point, and g(x)=h(x). You can get existence out of (say) the Hahn-Banach theorem (but you can also get it in an elementary way.)

The statement that L^1 is the completion of C_c in the norm given by the Riemann integral is just a rehash of the uniqueness of the completion of a metric space. We know that in the Lebesgue construction, C_c is dense in L^1 (and so L^1 is a concrete completion of C_c). If we define a norm on C_c using the Riemann integral, we know that it agrees with the norm you would get out of restricting the L^1 norm on C_c. Hence, C_c with the Riemann integral L^1 norm, is the same as C_c with the Lebesgue integral L^1 norm. C_c with the Riemann integral L^1 norm has an abstract completion, as per metric space theory, but the Lebesgue theory is just an explicit version of this completion.

Loisel 19:44, 27 Jun 2004 (UTC)

Defining the integral of simple functions[edit]

The article states that the integral is extended to simple functions by linearity. This is not strictly correct. Despite that fact that once properly defined on simple functions (and checked well defined), the integral can be proved to be linear. However, it has not yet been defined on simple functions, only on indicator functions. The definition must be extended and linearity proved before the linear property can be used.

The result does not need to be extended however for disjoint simple functions (those with disjoint sets), since it follows by considering the union and using the additive property of the measure. But it is not for general simple function where non-empty intersection is allowed. This does not follow from the definition thus far, and the definition must be extended to cover it. This can obviously be done by simply defining it to be what it would be if linearity were to hold (because it does, and can be proved as such). However, the key point I am making is that the proof that such a definition is well defined is not trivial (but doable in about half a page or less). And this is the key to the creation of a sound lebesgue integral. It is required in the proof to index all possible intersections of these sets. This essentially allows the problem to be broken down to a situation of disjoint sets and in this way we can complete the proof.

I think that even if a new page has to be created for the proof, it is fundamental to the sound definition of the lebesgue integral. I would be happy to write the proof out myself.

--Wikiphile1603 (talk) 00:32, 24 March 2009 (UTC)

"Wikipedia is not a textbook": writing half a page about the definition of the integral of simple functions is clearly in opposition with it. Point to the difficulty, explain the idea of solution and send to a book with chapter and page, all this in three lines, should be enough in my opinion. I disagree having a separate page for this tedious proof. --Bdmy (talk) 08:38, 24 March 2009 (UTC) --Bdmy (talk) 08:38, 24 March 2009 (UTC)
Thank you for your response. Actually I agree with you. It's easy to get a little carried away with trying to make this perfectly accurate. As you say, this level of rigor would make it like a textbook. I will consider if it is worth complicating the situation on the page at all, since without a fairly deep explanation I fear it may be confusing and people would miss the subtlety. --Wikiphile1603 (talk) 11:33, 24 March 2009 (UTC)

Indicators[edit]

Does anyone agree with me that there should be more consistency with the notation here? On the pages indicator function and Riemann integral, I(your set here) is used. I definitely prefer this, or a notation with χs, to the current one on this page. Best, mat_x 15:59, 18 Aug 2004 (UTC)

Mountains[edit]

The Riemann integral of the mountain requires lower sums and upper sums, in each vertical slice find the highest and lowest points. The integral beng described is more like the Cauchy integral. CSTAR 15:42, 19 Dec 2004 (UTC)

Towards a better integration theory[edit]

Is there any objection to deleting this section? CSTAR 17:42, 22 Dec 2004 (UTC)

I have no objections. You are doing a good job on the page, but I wonder whether you could write slightly more on the definition of measure besides refering the reader to the measure (mathematics) article. I fear that many readers may not get pass the sentence "Let μ be a (non-negative) measure on a sigma-algebra X of subsets of E." -- Jitse Niesen 21:28, 22 Dec 2004 (UTC)
Yes I do plan to fix that soon.CSTAR 21:35, 22 Dec 2004 (UTC)

Edits[edit]

Remarks on measurability condition.. MEasurability of a function as defined in the article is a property of the underlying sigma-algebras of source and target space. The user who made the latest edits, confused this with measurability of functions with respect to the sigma-algebra of the completion in the source space. This is not necessary and is nowehere used in the article.. Of course this means that if f = g a.e. and is measurable g may not be. There are a number of other changes which I don't see add anything to the article. For instance why add vector-valued functions here? If there is no objection I am going to revert. The only improvement was the use oof \liminf and \limsup whic I propose to keep. CSTAR

Integral as area under curve[edit]

Intuitively, continuity is not necessary: Two disjoint rectangles with vertical sides and of different heights have an area and aren't representedby continuous functions. CSTAR 15:29, 11 Mar 2005 (UTC)

The first sentence of the article talk about an area being bounded. If the fuction is discontinuous then no area is bounded. Of course I should also have made changes to say that the area is bounded from below by 0. I realize that even then you will have difficulty on the sides, so I will be bold and change it to something which mentions area below a (positive) function. MarSch 15:40, 11 Mar 2005 (UTC)
Huh? Discontinuity implies unboundedness? What about the indicator function of the interval [0, 1]? That's a bounded function with a discontinuity. Michael Hardy 00:09, 14 Mar 2005 (UTC)
Perhaps MarSch means with "unbounded" that the graph does not run along the boundary. For instance, with your indicator function, the graph looks like this
              ------



      --------      ------
and the vertical sides of the square — running from (0,1) to (0,0) and from (1,1) to (1,0) — are not bounded by the graph. MarSch's change from "bounded" to "contained" solves the problem. -- Jitse Niesen 14:12, 14 Mar 2005 (UTC)
Referring to the region under the curve and above the x-axis is about as unambiguous as you can get in natural language. And the anti 2D bias isn't very informative in my view. CSTAR 14:26, 14 Mar 2005 (UTC)
I think we should try to keep the first sentence as easy as possible and I edited the article accordingly (bringing back the "2D bias"). -- Jitse Niesen 15:10, 14 Mar 2005 (UTC)
As far as I've seen there are also articles about integration and Riemann integration. Your sentence would be perfect for integration. For those reading this article it is needlessly simplistic and ugly. -MarSch 14:40, 4 Apr 2005 (UTC)

Generalization[edit]

This article defines the integral for functions f: R -> R. From what I can tell from reading it, it's trivial to extend this definition to cover any function f: S -> R where S is any measurable set - should the article cover this? I'm particularly thinking that it's easy to define an integral on f: R x R -> R — ciphergoth 22:30, 2005 Apr 28 (UTC)

From what I see, the integral is defined exactly as you want, on a general space. The first several paragraphs deal with this. Oleg Alexandrov 23:08, 28 Apr 2005 (UTC)
Just on a style point, instead of writing f: R -> R you could write <i>f : R -> R</i> Wikipedia compiles it as f : R -> R. The you could replace the arrow -> with → . To get that arrow scroll down just under where it says "Do not copy text from other..." there's a scroll box. Choose "Insert" and there you can click the arrow. So this brings us to f : R → R. Finally <b>R</b> will be complied as R, so that if we write <i>f : <b>R</b> → <b>R</b></i> Wikipedia compiles it as f : RR. Looks better, no? Similar sytax will give you things like f : RmRn. But hey, maybe I just have too much time on my hands.  Declan Davis   (talk)  17:14, 1 October 2008 (UTC)

Limitations of the Riemann integral[edit]

You show that the two functions f(x)=-1+2[x>0] and g(x)=-1+2[x>1] are not riemann integrable, and that the cauchy values of the integrals differ. Sadly. But it is not clear to me if the lebesgue integral improves the situation ? Bo Jacoby 13:57, 22 September 2005 (UTC)

I'm confused by this: "Failure of monotone convergence. As shown above, the indicator function 1Q on the rationals is not Riemann integrable. In particular, the Monotone convergence theorem fails. To see why, let {ak} be an enumeration of all the rational numbers in [0,1] (they are countable so this can be done.) Then let

The function gk is zero everywhere except on a finite set of points,..." How are the rational numbers in [0,1] a finite set? Aren't there countable but infinite pairs of integers that would correspond to rational numbers in that range? Gluuuuue (talk) 09:35, 17 August 2010 (UTC)

Reversion[edit]

Hi CSTAR. You reverted my simplification. Your argument is incorrect. The function |f|, mapping x to |f(x)|, allows any real function to be split into a positive and negative part: f=(f+|f|)/2+(f-|f|)/2. The ad hoc notations f+ and f- are not necessary. Bo Jacoby 08:23, 27 September 2005 (UTC)

Reply.

  1. Your definition of the integral was expressed as a difference of two values both of which could be +∞ You must assume that at least one of the values is finite.
  2. Re your comment: The ad hoc notations f+ and f- are not necessary and should be avoided. On the contrary, this a very common notation and is hardly ad-hoc. See
  • P. Halmos, Measure Theory, or
  • W. Rudin, Real and Complex Analysis.
Why do you believe that this notation should be avoided?
--CSTAR 15:57, 27 September 2005 (UTC)
  1. Thank you for your reply.
  2. I agree that the finity condition should not be postponed.
  3. Using the better-known absolute value function |x| should make it easier to read. x+ = (x+|x|)/2 and x- = -(x-|x|)/2. You may assume that the reader knows |x|, but as you do not assume that the reader knows x+ and x-, you state that 'we need a few more definitions' and then you define x+ and x- in the text. Well, we don't need these definitions. We can use standard mathematical expressions instead.
  4. Please compare my version with your reverted version. You state that the integral is defined for real functions, but you actually define it for complex functions too. Why not state that it is defined for complex functions? You call the set X when you mean E. It's merely a typo. Your definition made in the 'indicator function' section was not used in the 'simple function' section , so either the former could be omitted or the latter be changed to build on the former. I made many small improvements.
  5. I wonder why the classical notation f(x)dx is abandoned. How do you express the integrand ax^2dx as opposed to ax^2da when you write ax^2d\mu? Only the repeated dummy variable tells what the integration is about.

Bo Jacoby 11:53, 28 September 2005 (UTC)

i second CTARS's comments above. these notations are standard, not "ad hoc". Mct mht 18:01, 16 August 2006 (UTC)

Silly quote[edit]

I removed a silly quote about the Lebesgue integral not being useful "physically". I don't know what this means. Is there a physical usefulness to having the irrationals? The reals are to the rationals as Lebesgue is to Riemann. And, should we add a ton of pro-Lebesgue quotes? -cj67

Re: And, should we add a ton of pro-Lebesgue quotes? I don't care one way or the other.
Re The reals are to the rationals as Lebesgue is to Riemann. I don't know what this means. More specifiuically, the reals are the metric completion of the rationals; the extension of Riemann to Lebesgue is more subtle than an extension by continuity to a completion.--CSTAR 17:16, 26 May 2006 (UTC)
Nope. Lebesgue integrable functions are the completion of the continuous Riemann integrable functions w.r.t. the L^1 norm (which can be defined using the Riemann integral, since you are only considering continuos functions).(Cj67 20:05, 30 May 2006 (UTC))
Nope. The elements have to be represented as functions. As abstract Banach spaces, L^1 is the completion of the continuous Riemann integrable functions w.r.t. the L^1 norm, as you point out. But representing an element of this Banach space as an equivalence class of measurable functions is another matter, because pointwise evaluation is not continuous as a functional on L^1. --CSTAR 20:24, 30 May 2006 (UTC)
L^1 is a representation of the completion of Riemann integrable functions, just as the reals are a representation of the completion of the rationals. I think the analogy is clear. Of course, there is more involved, since L^1 does not equal the reals. But I think my point about the physical "usefulness" of the irrationals is clear. (Cj67 22:15, 30 May 2006 (UTC))
the analogy looks fine to me. completion of metric spaces are by construction equivalence classes. the Lebesgue-a.e. condition gives an explicit identification. Mct mht 18:08, 16 August 2006 (UTC)
My point in saying that "the extension of Riemann to Lebesgue is more subtle than an extension by continuity to a completion" was that the "pointwise" properties of L^1 do not follow by continuity, since f_k converges to f in L^1 does not imply f_k converges to f almost a.e. (though it is true that some subsequence of f_k converges to f ae.) . ANyway, this is ome of those WP talk page discussions I'm sorry I ever got into.--CSTAR 18:22, 16 August 2006 (UTC)
  • A curious thing: ain't integration by parts allowed when performing a lebesgue integral under certain measure (i.e Gaussian measure), i don't know if this is omitted in the article or it simply can't be a Lebesgue integration by parts under a given measure.

extended real number line.[edit]

Article quote:

Let f be a non-negative measurable function on E which we allow to attain the value +∞, in other words, f takes values in the extended real number line.

"in other words" is not quite true, as the extended real number line also contains negative real numbers and minus infinity.

Bo Jacoby 13:46, 7 February 2007 (UTC).

It seems fine to me. We write f : RR to represent a function f such that for each real number x we get another real number f(x). There is no need for f to be surjective, i.e. for every real y there does not have to be a real x such that y = f(x). For example, the function f(x) := x2 is a real valued function, and we could write f : RR, although we could also write f : R → [0,∞) if we wanted to. The phrase "f takes values in the extended real number line" means that f(x) will be in the extended real number line. It does not mean that the image of f will be equal to the whole extended real number line.
 Declan Davis   (talk)  16:42, 1 October 2008 (UTC)

Thanks for the comment. The phrase "statement P, in other words statement Q" means "P \Leftrightarrow Q \,". So the sentence "Let f be a non-negative measurable function on E which we allow to attain the value +∞, in other words, f takes values in the extended real number line" means that "Let f be a non-negative measurable function on E which we allow to attain the value +∞"  \Leftrightarrow  \, "f takes values in the extended real number line", which is not true. Only the arrow " \Rightarrow  \," is true. The arrow " \Leftarrow  \," is false. The fact that f takes values from the extended real number line does not imply that f is a non-negative measurable function on E. Bo Jacoby (talk) 19:34, 1 October 2008 (UTC).

I see, you're quite right! Did you correct the article?
 Δεκλαν Δαφισ   (talk)  02:10, 2 October 2008 (UTC)

The 'intuitive' interpretation[edit]

Hi. The 'intuitive' interpretation for the Lebesgue integral is one I have seen before, but doesn't seem to correspond to what is actually going on. According to the mountains diagram for the Lebesgue integral we are multiplying increments in the value of f by the measure of the set of all x which map to equal or higher values of f. This is incorrect, in reality we multiply f (not it's increments) by the measure of all x which map to that value of f. I realize that the result is the same, but the process is different, and so the 'intuition' is a bit confusing when trying to understand the actual correct definition. For a more correct mountains diagram you could imagine a curve with upright rectangles filling it from below (like the dark blue rectangles for the Riemann integral) but where rectangles of different heights have different colours, and those of the same height have the same colour and correspond to one term of the summation for the integral of a simple function.

Benjaminveal 11:59, 3 June 2007 (UTC)

Benjaminveal is right; that section could use some reworking. also, in comparing with the Riemann integral, the following can be pointed out: the Riemann sum is obtained by partitioning the domain of a function. therefore the limit may not be well-behaved if the function exhibits violent local behavior on its domain, e.g discontinuous. this is rectified in the Lebesgue approach by instead partitioning the range of a function and consider the preimages, which necessitates the notion of a measure of a fairly arbitrary set, the family of which is called the Lebesgue-measurable sets. an indication of the effectiveness of the Lebesgue approach is that there's no constructive example of a non-Lebesgue measurable set.

also, perhaps i missed it but the article doesn't seem to explicitly mention that Lebesgue integral extends the Riemann integral and that the Riemann integral functions are classified via the Lebesgue measure. Mct mht 14:19, 4 June 2007 (UTC)

I have rephrased this section because I didn't think it was all that well written. I have also added a reference for those of you who are detractors. Loisel 21:59, 18 June 2007 (UTC)

to build the simple function corresponding to a given partition of the y axis, one passes back to the preimages and thus ends up with vertical-looking rectangles. if we insist on having a picture, wouldn't it be better, more precise, to reflect this? Mct mht 02:35, 19 June 2007 (UTC)
We had the same problem in the german wikipedia. If you want to, you can use the picture from there de:Lebesgue-Integral--91.23.26.165 (talk) 10:53, 21 September 2011 (UTC)
Thanks, I've replaced the red & blue figure with the one you suggest.--agr (talk) 11:09, 21 September 2011 (UTC)

Figure in red and blue[edit]

I think that the figure in red does not match with lebesgue integral definition (although blue one does)

--Nicolaennio 21:23, 9 September 2007 (UTC)

That's the same I think... But for some unknown reason everybody tries to explain the difference between lebesgue and riemann integral with these two pictures. I don't get it --188.103.126.158 (talk) 23:37, 3 August 2011 (UTC)

Faulty discussion of integration on unbounded intervals[edit]

The discussion of the faults of the improper Riemann integral seems faulty to me. For some reason, the only definition of improper integral considered is the symmetric limit (a.k.a. the Cauchy principal value), with the justification that it is the "simplest possible" definition. This definition is too simple, for the reasons described in the example: it fails translation invariance (or, as I try to explain to my calculus students, making a u-substitution can change the answer). But this is a strange argument: on the one hand, for these reasons this CPV-style limit is not the definition of an improperly Riemann integrable function given (even) in calculus books: rather, one picks a real number a, breaks up into two improper integrals and requires that both integrals converge. (There are of course more elegant ways to express the same thing, e.g. by convergence of a net based on the collection of all closed subintervals of the real line ordered under containment.) With the standard definition, the given function is not improperly Riemann integrable. Moreover, it is not Lebesgue integrable either, so this example is quite misleading. Indeed, one of the features of the Lebesgue integral (whether or not it is a drawback depends on your perspective) is that it does not include all improperly Riemann integrable functions.

I will give the writers of this otherwise very nice article a chance to correct this before I remove it. 70.155.76.162 (talk) 02:15, 27 November 2007 (UTC)Plclark

Mix between the general Lebesgue integral and the specific one that extends the Riemann integral.[edit]

At some places in the article there is a mixture between the two meanings of Lebesgue integration. To quote from the article: The term "Lebesgue integration" may refer either to the general theory of integration of a function with respect to a general measure, as introduced by Lebesgue, or to the specific case of integration of a function defined on a sub-domain of the real line with respect to Lebesgue measure.

It is not a bad thing to use the second meaning troughout the article as a guidline and/or example but it should be clear in every statement which meaning is used. I would do it myself but I don't feel proficient enough (yet!) with Lebesgue integration at this point.

Stigin (talk) 16:04, 16 January 2008 (UTC)

Lebesgue integration is almost universally understood to mean the general theory of integration with respect to a countably additive measure. You're right though that this made need clarification in the article.--CSTAR (talk) 16:19, 16 January 2008 (UTC)

Minor Change (Possibly)[edit]

The Riemann integral is the sum of rectangles as width goes to zero, so wouldn't "Riemann's definition starts with the construction of a sequence of easily-calculated areas (as opposed to integrals) which converge to the integral of a given function" be the correct way of putting it? I made the correction but Im an undergrad so please feel free to slap me. See Here —Preceding unsigned comment added by Dustankb (talkcontribs) 19:44, 1 October 2008 (UTC) --Dustankb (talk) 15:59, 1 October 2008 (UTC)

Could you please leave a link to the section in question? If you want to link to the section "Proof Techniques" and you wank it to appear as a blue link saying "See Here" then you type [[Lebesgue_integration#Proof_techniques|See Here]] and Wikipedia will then make it come out as See Here.  Declan Davis   (talk)  16:18, 1 October 2008 (UTC)
The "See Here" in my original statement has been corrected to jump to the point in question, I made the change some time ago, and with no correction since then I guess its safe to say it was a valid correction. If this however is not the case please contact me/correct as needed.

Chinese Version[edit]

The 12:02, 31 August 2008 Ht686rg90 version of this article is translated into Chinese Wikipedia.--Wing (talk) 18:39, 12 September 2008 (UTC)

L1 definition[edit]

There is no definition, or link to a definition, given for a function to be Lk integrable. Even though lines like "...since f is L1 integrable..." turn up in proofs. This has never been my subject area. I seem to have a recolection of

 \int_E |f(x)|^k \ d\mu_E

being involved somewhere. Could someone please add the necessary link, and or definition. Maybe an article could be written on the very subject?  Δεκλαν Δαφισ   (talk)  16:28, 2 October 2008 (UTC)

Lp space?Ht686rg90 (talk) 16:36, 2 October 2008 (UTC)
Super. Have you added the links?  Δεκλαν Δαφισ   (talk)  17:00, 2 October 2008 (UTC)
Seems not. Well I have just added it.  Δεκλαν Δαφισ   (talk)  17:08, 2 October 2008 (UTC)

pronunciation[edit]

An editor removed the pronunciation of Henri Lebesgue because the article is on the integral and not its inventor and because it was in Henri's article. I can see that, however, I'd like to know how to pronounce the integral (or at least how most mathematicians pronounce it, I'd like to sound like I know what I'm tlaking about). RJFJR (talk) 04:08, 10 November 2008 (UTC)

Integration of simple functions[edit]

The definition in this page is not correct, as the measures of the sets could be infinite. What is the value of the following integral of a simple function

 \int \bigl( \mathbf{1}_{[0, \infty)} - \mathbf{1}_{(-\infty, 0)} \bigr) \, \mathrm{d}\lambda,

where λ is the Lebesgue measure? Bdmy (talk) 07:58, 5 December 2008 (UTC)

What is the difference?[edit]

What is the difference between the definition given in the article and the one given below (which seems much more intuitive to me)? Is the one in the article more general? Is my definition not restrictive enough? Is there some massive gap in my understanding? etc.

My approach is simply to define the integral as the area/volume/hypervolume "under the curve". Since that's exactly what we think of the integral as being, it would seem to me that this definition should be a quite natural approach.

Let E be a measurable subset of Rn. Let μ be a Lebesgue measure on Rn+1. Let f be the function for which we will attempt to find an integral over E (domain includes E, range includes R, just like the article).

For non-negative f:

 \int_E f d \mu = \mu(\left\{(x_1, x_2, x_3... x_n, y):0 \le y \le f(x_1, x_2, x_3... x_n) \and (x_1, x_2, x_3... x_n) \in E\right\})

If the expression on the right-hand-side is undefined, then the integral is undefined.

For signed f: similar to the article approach. When ∞ - ∞ is encountered, the integral is undefined.

Equivalently, a succinct definition for all f:

 \int_E f d \mu = \mu(\left\{(x_1, x_2, x_3... x_n, y):0 \le y \le f(x_1, x_2, x_3... x_n)\and (x_1, x_2, x_3... x_n) \in E\right\}) - \mu(\left\{(x_1, x_2, x_3... x_n, y):f(x_1, x_2, x_3... x_n) \le y \le 0\and (x_1, x_2, x_3... x_n) \in E\right\})

All responses appreciated. --69.91.95.139 (talk) 02:20, 23 January 2009 (UTC)

As far as I understand, what you say is correct, and follows from the Fubini-Tonelli theorem, once the theory is established, but I would insist to distinguish in the formulas the measure \mu_n on Rn from \mu_{n+1} on Rn+1. The problems with your approach are perhaps that:
  • when integrating on the line you already need to know the (more complex?) two-dimensional case; even worse, when integrating over some abstract space Ω, you need to know from the start about the tensor product with the Lebesgue measure on the line, and I think that it is much more comfortable to deal with the construction of the tensor product after the integral is constructed.
  • simple facts like linearity of the integral may be harder to see
Once, long ago, I tried to teach Lebesgue integral by saying: don't be affraid, this is just a Riemann integral (for ƒ ≥ 0), by
\int f \, d\mu = \int_0^\infty \mu( \{ \omega : f(\omega) > t \}) \, dt
but I stopped doing it after one or two years, because of the second point above, and also because it was hard for students to make the link with the other courses that had different viewpoint.
--Bdmy (talk) 08:30, 23 January 2009 (UTC)
Well, ignoring the fact that I forgot to assert that (x_1, x_2, x_3... x_n) \in E (fixed now). Anyway, now I feel like I have a better understanding of the approach given in the article. I still don't feel like I fully understand the motivation behind it, but I'm not going to argue the point when I don't have the background to do so.
Thanks for your help. --69.91.95.139 (talk) 14:59, 24 January 2009 (UTC)
About the set E, it is just as simple to forget completely about it (let's pretend it is coded in the set where ƒ ≠ 0), and just say for ƒ ≥ 0 and measurable:
 \int f d \mu_n = \mu_{n+1}(\left\{(x_1, x_2, x_3... x_n, y):0 \le y \le f(x_1, x_2, x_3... x_n) \right\})
Now try for yourself to take n = 1 and to compute the integral of ƒ+ g; you will find that you need properties for the two-dimensional measure, reasonable but not proved, as this one: the measure of a subset of R2 does not change if you translate independently (in a measurable manner) every vertical section of the set.
I agree though that "your" formula could be reassuring for the reader, and should perhaps be included in the article. --Bdmy (talk) 15:33, 24 January 2009 (UTC)

Writing style[edit]

We should clean up the writing style in this article. I cleaned up the lead to make it more direct and less conversational. There is a lot more to do. Sentences like, This is easy to understand for familiar functions such as polynomials, but what does it mean for more exotic functions? are not encyclopedic. Is it really "easy to understand"? What does "easy to understand" mean, and why is that even important? Also, instead of asking the question about what it means for more exotic functions, we should just say that it has no meaning, or it is not well defined.

That's just one example. There's some guidance given at WP:Manual_of_Style_(mathematics). I followed its advice to move the names, dates, related fields, etc. into the first couple of paragraphs, and left the informal introduction to the topic for the next couple of paragraph.

The use of the word "we" is discouraged in the article, and I think it's possible in this article to avoid it. It will force us to write more directly. Sancho 05:35, 20 February 2009 (UTC)

A request for an interesting example, pretty please[edit]

Hi!

Could someone find it in his or her heart to give an example of a Lebesgue- but not Riemann-integrable (proper or improper) function other than 1_Q? A "reasonable" function would be ideal, but then reasonable functions are already Riemann integrable. But just one example of a "non-reasonable" function which might actually pop up in "real life"? Or failing that, just any interesting example? The example 1_Q is the only example one ever sees in textbooks and it is not really impressive. One can even guess the value knowing almost nothing about the theory. All the best 157.157.28.110 (talk) 19:11, 9 February 2010 (UTC)

PS: From the article:
For example, the Dirichlet function, which is 0 where its argument is irrational and 1 otherwise, has a Lebesgue integral, but it does not have a Riemann integral.
This "for example" has been seen too often to count. I'll add a question: Have other examples been published elsewhere at all? If so, I'd be grateful to see them. Cheers 157.157.28.110 (talk) 19:28, 9 February 2010 (UTC)

I would also be interested in such an example - especially in an example which is not derived from a Riemann-integrable function by just changing the function's value on a set of measure zero. Is that possible?

In other words: Is it possible for every Lebesgue-integrable function to find a Riemann-integrable function by just changing the function’s value on a set of measure zero?

Background: The standard example of a Lebesgue-integrable function which is not Riemann-integrable is the indicator function of Q or the Cantor Set or some other set of measure zero. Now I am asking myself: Is it possible to construct a real Lebesgue-integrable function which is not just a Riemann-integrable function changed on a set of measure zero? 109.84.0.136 (talk) 21:05, 14 November 2011 (UTC)

I have heard that Lebesque integration is used in stock market analysis. I can also see it being useful in fractal geometry. I am not an expert here, though, unfortunately. I am a physicist by training and I was hoping to learn some information about this, but translating from mathese to physicsspeak takes way too much time even when the mathematicians care about the same issues as the physicist. (They often don't.) The stuff I want to see like why it is used and how it is used and what it is used for is buried deep in the article. At least it seems to be mostly there, though. TStein (talk) 21:16, 10 August 2010 (UTC)

Some errors in the figure demonstrating the interpretation of Lebesgue integration.[edit]

I think there is an error in the figure demonstrating the difference between Lebesgue integration and Riemann integration! The intuitive interpretation of Lebesgue integration is wrong!

When f is a non-negative measurable function, we define

\int_E f \, \mathrm{d}\mu = \sup\left\{\,\int_E s\, \mathrm{d}\mu : 0 \le s \le f,\,\right\},

where s is a simple function. If we divide the range of f into n intervals equally, let $$I_{n,k}$$ be the k-th interval, and define the measurable sets $$A_{n,k} = f^{−1}(I_{n,k})$$. Then

s_n = \sum_k a_k 1_{A_{n,k}}.

The problem is where is the measurable set $$A_{n,k}$$ in the figure?

The figure in question has been replaced.--agr (talk) 15:24, 21 September 2011 (UTC)
The original figure has been restored, please see 1 and 2. --151.75.35.106 (talk) 09:48, 15 March 2012 (UTC)
Further discussion is going on here: Wikipedia_talk:WikiProject_Mathematics#Lebesgue_integration.--Svebert (talk) 09:59, 19 March 2012 (UTC)

The definition has disappeared[edit]

In the attempt to make things simpler, the article doesn´t even state what Lebesgue integrable means. Or at least, it is not clearly stated. Maybe it is somewhere hidden in the middle of the text. --190.188.2.122 (talk) 15:12, 6 June 2011 (UTC)

Dubious?[edit]

I recently removed the two tags that were just added. It is, of course, commonly acknowledged, either explicitly or implicitly, that the Lebesgue integral partitions the range of the function rather than the domain. This is supported by the reference to Lebesgue himself, the reference to Folland, and (now) the reference to Lieb and Loss. So, at least as far as I see, the statement in the article is not only not wrong, but it is supported by the surrounding text and the references contained therein.

However, in case I am wrong about some matter of interpretation, it is usually customary when adding (or in this case removing) tags such as these to include a discussion on the talk page. The issue of whether this interpretation is "wrong" has been discussed (and in my opinion adequately rebutted) here. Sławomir Biały (talk) 02:03, 15 September 2012 (UTC)

No notion of improper Lebesgue integral?[edit]

I disagree with this edit saying that there is no notion of improper Lebesgue integral. For example, if f\in L^2(\mathbb R), then the Fourier transform of f is defined by the improper Lebesgue integral

\hat{f}(\xi) = \int_{-\infty}^\infty e^{-2\pi i x\xi}f(x)\,dx.

It's only when f\in L^1(\mathbb R) that this makes sense as a proper Lebesgue integral. Sławomir Biały (talk) 12:18, 20 April 2013 (UTC)

Well I thought that the sentence referred to the non-existence of the notion of the value of an improper Lebesgue integral for that specific function, and I did not see how the value of its Riemann integral could be viewed in terms of Lebesgue integration. Perhaps it was not phrased well at all. Thanks for clarifying your interpretation, and I'll be glad to have you edit or remove it, whichever you deem suitable. Lim Wei Quan (talk) 03:03, 22 April 2013 (UTC)
Of course “improper integrals” can be introduced for any notion of integrals on the real line. Being able to integrate \sin(x)/x is no specific feature of the (improper) Riemann integral, it works as well when using for example the Lebesgue integral. You just need a notion of an integral such that you can integrate continuous functions on compact intervals. Then you can introduce a corresponding “improper integral” as a limit where the end-points approach infinity. In that sense every function that is improper Riemann integrable is also improper Lebesgue integrable (but not necessarily Lebesgue integrable). The concept of improper Lebesgue integrals is seldom used (contrary to the improper Riemann integral improper Lebesgue integrals are not typically mentioned in calculus text books—but are improper Riemann integrals used at all? I do not know any insightful usage of Riemann integrals). But 1. it exists in the mathematical literature (just use a search engine[1], and Sławomir Biały gave a good example (it is an approach to defining the Fourier transform on L^2(\R) (you can also just define it on the dense subspace L^1(\R)\cap L^2(\R) and extend it continuously to L^2(\R)—you instantly see that the definitions work)) 2. it should not sound like “in some respects Riemann integration is more powerful than Lebesgue integration”—it is not. --Chricho ∀ (talk) 14:51, 23 April 2013 (UTC)
Yes I know that we could extend by taking limits, but I thought one advantage of the Lebesgue integral is that it avoids problems with limits involving Riemann integrals. Also, I wish to respectfully disagree that the original sentence "sounded like Riemann integration is more powerful than Lebesgue integration". I personally view the Lebesgue integral as a more complete approach but I also feel that having improper integrals on unbounded intervals is a more natural extension for the Riemann integral. Also, I just found a similar statement at http://en.wikipedia.org/wiki/Improper_integral#Types_of_integrals. Should there be a link to there for further information? Thanks! Lim Wei Quan (talk) 04:36, 25 April 2013 (UTC)
In the Fourier transform example, the improper integral does not exist in general. So one is not "extending by taking limits" in the sense of integrals. When the improper integral does exist, improper Lebesgue integral extends the Riemann integral just by definition.
Lebesgue integral is better behaved under pointwise limits than Riemann, but you still need either monotone convergence or integrability(L1-dominance). Absent those assumptions, one still needs to resort to improper integration. Mct mht (talk) 05:21, 25 April 2013 (UTC)
Not disputing the point being made but the Fourier transform is not really an honest example, though. It's defining a unitary operator by extending from a dense set in L^2(\R), while the the right hand side may or may not exist as an (improper Lebesgue) integral. Mct mht (talk) 08:15, 24 April 2013 (UTC)
The description as an improper Lebesgue integral is just a little bit less abstract and a valid viewpoint. Of course this does not show a “profound meaning” of improper Lebesgue integration (I guess there is no profound meaning, just like there is not anyone for improper Riemann integration), it does not really matter whether you choose this definition or say “extend from a dense subspace”, but this viewpoint might help to get some intuition. --Chricho ∀ (talk) 21:23, 24 April 2013 (UTC)
As Chricho says, this is a valid viewpoint. Take your dense subspace (of L^1\cap L^2) to consist of functions of the form f\chi_{[a,b]} where f\in L^2(\mathbb R). The right hand side exists as an improper Lebesgue integral for f\in L^2 (at least in the L^2 sense). In fact, this is how the Fourier transform in L^2 was classically defined by old-school harmonic analysis like F. Riesz and Titchmarsh. Of course, the modern way is to define it by extending by continuity from L^1\cap L^2, which is generally preferred since it allows different basis functions (notably Gaussians). Sławomir Biały (talk) 12:31, 25 April 2013 (UTC)

Great explanation[edit]

One of the best math pages I've read on WP. We should nominate it for some kind of award. Crasshopper (talk) 15:51, 16 January 2014 (UTC)

Understanding the "Towards a formal definition" Section[edit]

The "Towards a formal definition" Section says that the area of the horizontal strip between y=t and y=t+dt is given by \mu \left (\{x\mid f(x)>t\} \right ) \,dt.. I don't see why this is the case at all. I can understand why we're multiplying by dt, because that's the width of the strip, but I don't understand where we're getting the length of the strip.